$\begin{aligned} y&=(-x^2+6x-4)^5 \\\\ \dfrac{dy}{dx}&=\,? \end{aligned}$ Choose 1 answer: Choose 1 answer: (Choice A) A $5(-2x+6)^4(-x^2+6x-4)$ (Choice B) B $5(-2x+6)^4$ (Choice C) C $5(-x^2+6x-4)^4(-2x+6)$ (Choice D) D $-10x^4(x^5-3)$
Since $(-x^2+6x-4)^5$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. $\underbrace{(~\overbrace{-x^2+6x-4}^{\text{inner}}~)^5}_{\text{outer}}$ So if $(-x^2+6x-4)^5=w(u(x))$, then: $\begin{aligned} {u(x)}&={-x^2+6x-4} &&\text{inner function} \\\\ w(x)&=x^5&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={-2x+6} \\\\ {w'(x)}&={5x^4} \end{aligned}$ $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={5({-x^2+6x-4})^4} \cdot {(-2x+6)} \end{aligned}$